To find the intersection between a straight line and a cubic piece of a spline, write the line as Ax+By+C=0, then substitute x and y by their values as a function of the parameter t. You should get a cubic equation on t, which you can either solve in terms of radicals (tricky) or using Newton's method (easy).

If you have trouble with that plan, explain how far you got and what part you can't get to work.

I've been able to google a lot of information i wasn't aware of before thanks to your explanation.

To clarify.

Using Newton's method to find the root of (Ax+By+C=0), where X and Y is a point on the curve segment.Taking an initial guess for X ,Y (i presume t = 0.5)Subsequent iterations narrow down the accuracy of the curve point X , Y until the root is found.

In code terms, a quick google search provided (Ax+By+C=0) to be

(y1 – y2)x + (x2 – x1)y + (x1y2 – x2y1) = 0

If my above assumptions are correct, all thats needed now is to write Newtons Method to find the root.

I think you are not doing it the way I described, so maybe I'll describe it in more detail.

A cubic spline is piece-wise cubic, which means that it consists of several pieces of parametric curve (each for a range of values of t) that look like this:
x = a3*t^3+a2*t^2+a1*t+a0
y = b3*t^3+b2*t^2+b1*t+b0

If you now have an equation Ax+By+C=0, you can substitute the values of x and y from the parametrization above and you'll get
(A*a3+B*b3)*t^3 + (A*a2+B*b2)*t^2 + (A*a1+B*b1)*t + A*a0+B*b0 = 0

Now the equation is C3*t^3 + C2*t^2 + C1*t + C0 = 0. Plug this into a solver that can get you all the real roots and look for roots in the range of values of t that this piece is for. The place where I was suggesting to use Newton's method was in the solver.